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Line integral across a field given by circular distribution

  1. Aug 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Evaluate [itex]\int_C \vec{F} \cdot d\vec{r}[/itex]

    Where [itex]\vec{F}[/itex] is the field generated from a circular thread of radius b in the xy plane, with magnitude j in the direction [itex]\hat{\varphi}[/itex] (i.e. not along the curve, I take it)

    C: [itex](x,y,z) = b(1+ \cos{\alpha}, 0, \sin{\alpha})[/itex]

    3. The attempt at a solution


    I know that I probably could use Stoke and get -j (or a 0) as an answer here, due to the geometry, but I want to practice using the naïve approach and compute the integral, the hard way, directly. But, the fact that phi describes a rotational field around the line, and not along it, is a bit fishy.

    To begin with, I need to find an expression for [itex]\vec{F} [/itex].

    Here, I hesitate. [itex] \vec{F}= \int^{2\pi}_0 \frac{j \hat{\varphi}}{2\pi \cdot |\vec{r}-\vec{r}'|} d \gamma' [/itex] with phi relative to the source [itex]\vec{r}'[/itex], i.e. the circular thread. [itex]\vec{r}'=b \cos\gamma' \hat{x} + b \sin\gamma' \hat{y}[/itex]

    (Allowing for j as a function of angle would also be interesting in terms of implications to the potential use of stoke.)

    The next step would be to enter this integral into the main integral and integrate over alpha. However, I would eventually be bothered by [itex] \hat{\varphi}[/itex] which I need to convert to cartesian. The coordinate system of the thread, as I imagine it, is cylindrical, with height-z effectively taking the role of an angle. I would get a cartesian-x,y starting contribution from cosz and sinz and then r along that direction with respect to a function of phi. Then get cartesian-z from r and phi. Then I compute [itex] \hat{\varphi}[/itex] by differentiating the cartesian representation by [itex]\varphi[/itex]?

    Have I got the steps right? Any thoughts? Is it doable? Am I on the right track in that case?
     
  2. jcsd
  3. Aug 16, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Aug 18, 2016 #3

    BvU

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    Hello S,

    You are really making life difficult for yourself ! My respect :smile:

    I don't think your expression for##\ \vec r'\ ## is right. Check it in a few places.
    [edit]correction: this is the position on the current loop. Is OK.

    Here's (on page 9-35) a treatment on the off-axis field
     
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